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  • Using the method of integration, find the area of the region bounded by the lines <img alt="3x-2y+1= 0,2x+3y-21= 0\; and\; x-5y+9= 0" src="https://entrancecorner.codecogs.com/gif.latex?3x-2y&plus;1%3D%200%2C2x&plus;3y-21%3D%200%5C%3B%20and

Using the method of integration, find the area of the region bounded by the lines 3x-2y+1= 0,2x+3y-21= 0\; and\; x-5y+9= 0.

 

 

 

 
 
 
 
 

Answers (1)

Let  AB:3x-2y+1= 0
       BC:2x+3y-21= 0\; and       
       CA:x-5y+9= 0
ie:  AB:y= \frac{3x+1}{2}
BC:y= \frac{21-2x}{3}\; and
CA:y= \frac{x+9}{5}

clearly A\left ( 1,2 \right ),B\left ( 3,5 \right )\; and \; C\left ( 6,3 \right )
consider the diagram shown
\therefore Required\; area= \int_{1}^{3}y_{AB}\, dx+\int_{3}^{6}y_{BC}\, dx-\int_{1}^{6}y_{CA}\, dx
= \int_{1}^{3}\frac{3x+1}{2}dx+ \int_{3}^{6}\frac{21-2x}{3}dx- \int_{1}^{6}\frac{x+9}{5}dx
= \frac{1}{2}\left [ \left ( \frac{3}{2}x^2+x \right ) \right ]^{3}_{1}+\frac{1}{3}\left [ {\left ( 21x-x^{2} \right )} \right ]^{6}_{3}-\frac{1}{5}\left [ \frac{\left ( x \right )^{2}}{2} +9x \right ]^{6}_{1}
=7+12\: -\frac{25}{2}= \frac{13}{2}sq.units

Posted by

Ravindra Pindel

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